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Error when following answer concatenating csv files

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0

Hi i'm trying to follow this example, but receive the following error:

WindowsError: [Error 183] Cannot create a file when that file already exists

I've got a directory of 59 files labelled "Scan0" to "Scan58". I want to concatenate them whilst keeping the header of the first file.

I start by renaming the files into a common format:

import os
path = 'F:/ScanData'
i = 0
for filename in os.listdir(path):
 os.rename(os.path.join(path,filename), 
 os.path.join(path,'Scan'+str(i)+'.csv'))
 i = i +1 

Then using the answer referenced above I try and execute the concatenating code on the same directory:

fout=open("out.csv","a")
# first file:
for line in open("Scan0.csv"):
 fout.write(line)
# now the rest: 
for num in range(2,59):
 f = open("Scan"+str(num)+".csv")
 f.next() # skip the header
 for line in f:
 fout.write(line)
 f.close() # not really needed
fout.close() 

Yet this gives me the windows error. Any suggestions? I'm on a managed PC so cannot access shell or a terminal, else I would do this with sed or awk.

python-2.7 concatenation

share|improve this question

asked Mar 26 at 10:48

85567328556732

3366 bronze badges

add a comment | 

0

Hi i'm trying to follow this example, but receive the following error:

WindowsError: [Error 183] Cannot create a file when that file already exists

I've got a directory of 59 files labelled "Scan0" to "Scan58". I want to concatenate them whilst keeping the header of the first file.

I start by renaming the files into a common format:

import os
path = 'F:/ScanData'
i = 0
for filename in os.listdir(path):
 os.rename(os.path.join(path,filename), 
 os.path.join(path,'Scan'+str(i)+'.csv'))
 i = i +1 

Then using the answer referenced above I try and execute the concatenating code on the same directory:

fout=open("out.csv","a")
# first file:
for line in open("Scan0.csv"):
 fout.write(line)
# now the rest: 
for num in range(2,59):
 f = open("Scan"+str(num)+".csv")
 f.next() # skip the header
 for line in f:
 fout.write(line)
 f.close() # not really needed
fout.close() 

Yet this gives me the windows error. Any suggestions? I'm on a managed PC so cannot access shell or a terminal, else I would do this with sed or awk.

python-2.7 concatenation

share|improve this question

asked Mar 26 at 10:48

85567328556732

3366 bronze badges

add a comment | 

Hi i'm trying to follow this example, but receive the following error:

WindowsError: [Error 183] Cannot create a file when that file already exists

I've got a directory of 59 files labelled "Scan0" to "Scan58". I want to concatenate them whilst keeping the header of the first file.

I start by renaming the files into a common format:

import os
path = 'F:/ScanData'
i = 0
for filename in os.listdir(path):
 os.rename(os.path.join(path,filename), 
 os.path.join(path,'Scan'+str(i)+'.csv'))
 i = i +1 

Then using the answer referenced above I try and execute the concatenating code on the same directory:

fout=open("out.csv","a")
# first file:
for line in open("Scan0.csv"):
 fout.write(line)
# now the rest: 
for num in range(2,59):
 f = open("Scan"+str(num)+".csv")
 f.next() # skip the header
 for line in f:
 fout.write(line)
 f.close() # not really needed
fout.close() 

Yet this gives me the windows error. Any suggestions? I'm on a managed PC so cannot access shell or a terminal, else I would do this with sed or awk.

python-2.7 concatenation

share|improve this question

asked Mar 26 at 10:48

85567328556732

3366 bronze badges

WindowsError: [Error 183] Cannot create a file when that file already exists

import os
path = 'F:/ScanData'
i = 0
for filename in os.listdir(path):
 os.rename(os.path.join(path,filename), 
 os.path.join(path,'Scan'+str(i)+'.csv'))
 i = i +1 

fout=open("out.csv","a")
# first file:
for line in open("Scan0.csv"):
 fout.write(line)
# now the rest: 
for num in range(2,59):
 f = open("Scan"+str(num)+".csv")
 f.next() # skip the header
 for line in f:
 fout.write(line)
 f.close() # not really needed
fout.close() 

share|improve this question

asked Mar 26 at 10:48

85567328556732

3366 bronze badges

share|improve this question

asked Mar 26 at 10:48

85567328556732

3366 bronze badges

asked Mar 26 at 10:48

85567328556732

3366 bronze badges

asked Mar 26 at 10:48

85567328556732

3366 bronze badges

1 Answer
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0

Urgh I'm an idiot - I needed to remove the first bit of code for renaming the files, and then it works! Once I comment out the rename part the bottom code works fine.

share|improve this answer

answered Mar 26 at 12:03

85567328556732

3366 bronze badges

add a comment | 

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0

Urgh I'm an idiot - I needed to remove the first bit of code for renaming the files, and then it works! Once I comment out the rename part the bottom code works fine.

share|improve this answer

answered Mar 26 at 12:03

85567328556732

3366 bronze badges

add a comment | 

0

Urgh I'm an idiot - I needed to remove the first bit of code for renaming the files, and then it works! Once I comment out the rename part the bottom code works fine.

share|improve this answer

answered Mar 26 at 12:03

85567328556732

3366 bronze badges

add a comment | 

Urgh I'm an idiot - I needed to remove the first bit of code for renaming the files, and then it works! Once I comment out the rename part the bottom code works fine.

share|improve this answer

answered Mar 26 at 12:03

85567328556732

3366 bronze badges

share|improve this answer

answered Mar 26 at 12:03

85567328556732

3366 bronze badges

answered Mar 26 at 12:03

85567328556732

3366 bronze badges

answered Mar 26 at 12:03

85567328556732

3366 bronze badges